## Solve diophantine equation of form ax^2 + bx + c is a perfect square [closed]

Hello,

How do you solve diophantine equations so that ax^2 + bx + c is a perfect square

Thank you,

nonono

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What does this mean? Are you trying to solve $ax^2 + b x + c = y^2,$ for $a, b, c, x, y$ integers? – Igor Rivin Dec 31 2011 at 13:57
I assume that a, b and c are given and you want to find solutions x and y in integers. This can get complicated. If you want to find rational solutions, they exist if and only if there is no local obstruction. (This is the Hasse-Minkowski theorem.) Finding out whether there are integer solutions when you know rational ones exist involves computing in the class group of $\mathbb{Q}(\sqrt{a})$, and can be complicated. – David Speyer Dec 31 2011 at 14:45
I would suggest asking this over on math.stackexchange.com . I would really like to see someone write up an exposition of the quick blurb I gave you above, at a level that would be useful to someone who knows no advanced number theory, but many of the users here seem hostile to questions at that level. – David Speyer Dec 31 2011 at 14:47
This can indeed get complicated in general, but many important special cases (notably $b=0$ and $c=\pm 1$, or more generally $b=0$ and $|c|$ small) can be done without the class-group machinery — indeed such Diophantine equations were originally (and still can be used as) a major motivator in the development of this theory, and still contain nontrivial theoretical and computational challneges. A full exposition of this story is probably near the level but well beyond the scope of Mathoverflow, but if such an exposition already exists a pointer could be a very useful answer, esp. if annotated. – Noam D. Elkies Dec 31 2011 at 16:00
Firstly, I voted to close this, because the OP has difficulty with self-expression. I am happy to reopen it (indeed, I voted to reopen just now). Secondly, a propos Noam's comment: There is a bunch of stuff on this on the net: 1. alpertron.com.ar/QUAD.HTM actually solves equations. 2. This link: maths.ashwyninnovations.com/BQDE.pdf attributes the discussion to Gauss (Disc. Arith.) who, in turn, attributes the solution to Lagrange. It is quite detailed (and I hope correct, I didn't read in detail). If the question is reopened, I would be happy to post this as an answer. – Igor Rivin Dec 31 2011 at 19:26